描述

給定一個(gè)二叉樹，找出所有路徑中各節(jié)點(diǎn)相加總和等于給定目標(biāo)值的路徑。 一個(gè)有效的路徑，指的是從根節(jié)點(diǎn)到葉節(jié)點(diǎn)的路徑

樣例

給定一個(gè)二叉樹，和 目標(biāo)值 = 5:

             1
            / \
           2   4
          / \
         2   3
返回：

        [
          [1, 2, 2],
          [1, 4]
        ]

思路

遍歷二叉樹，將當(dāng)前結(jié)點(diǎn)值添加到路徑，遍歷到葉子結(jié)點(diǎn)得到每條完整路徑，判斷路徑中各點(diǎn)值是否滿足 target，將符合條件路徑添加到 result

代碼

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root the root of binary tree
     * @param target an integer
     * @return all valid paths
     */
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        ArrayList<Integer> path = new ArrayList<>();
        path.add(root.val);
        helper(root, path, root.val, target, result);
        return result;
    }
    
    private void helper(TreeNode root,
                        ArrayList<Integer> path,
                        int sum,
                        int target,
                        List<List<Integer>> result) {
        if (root.left == null && root.right == null) {
            if (sum == target) {
                result.add(new ArrayList<Integer>(path));
                return;
            }
        }
        
        if (root.left != null) {
            path.add(root.left.val);
            helper(root.left, path, root.left.val + sum, target, result);
            path.remove(path.size() - 1);
        }
        
        if (root.right != null) {
            path.add(root.right.val);
            helper(root.right, path, root.right.val + sum, target, result);
            path.remove(path.size() - 1);
        }
    }
}